help with two basic probability problems.

• Nov 25th 2009, 03:42 PM
santiagos11
help with two basic probability problems.
Problem 1:
There are 12 balls in an urn, 4 of which are white and 8 black. Three blndfold players, A, B, C draw a ball in turn, first A, thn B, then C. The winner is the one who first draws a white ball. Assuming that each (black) ball isreplaced after being drawn, find the ratio of the chances of the three players.

Problem 2:
There are 40 cards, 10 of each suit. A wagers B that he will draw four cards ad get one of each suit. What are the fair amount of wagers of each?

For problem 2, I know the answer (from the back of the book) is A:1000 vs. B:8139. which implies that the probability P = 1000/(1000+8139)=1000/9139. Also the number of combiantions of 4 cards out of 40 is
40 C 4 = 91390. Then the question that remains is how do i find (out of these 91390 different combinations) that there are 10^4=10000 combiantions that contain each card of a different suit?
• Nov 25th 2009, 04:23 PM
qmech
Suggestions
1. Until someone wins, isn't the probability on any turn 4/12 that that person will win (because the black balls are replaced)? If so, then the probability anyone will win is the probability that he'll win on that turn and that everyone has lost before.

2. A brute force way to figure out that probability is to consider the probability that you'll have drawn 2 cards from the same suit, or 3 cards, or 4 cards and subtract from the # ways of drawing 4 cards in general.
• Nov 25th 2009, 04:42 PM
santiagos11
Quote:

Originally Posted by qmech
1. Until someone wins, isn't the probability on any turn 4/12 that that person will win (because the black balls are replaced)? If so, then the probability anyone will win is the probability that he'll win on that turn and that everyone has lost before.

2. A brute force way to figure out that probability is to consider the probability that you'll have drawn 2 cards from the same suit, or 3 cards, or 4 cards and subtract from the # ways of drawing 4 cards in general.

on problem 2, the probability of drawing 4 cards of the same suit is (10/40)(9/39)(8/38)(7/37) = 21/9139. The probability of getting the complement (that is not getting 4 cards of the same suit) is (9139-21)/9139 or 1-(21/9139) which is equal to 9118/9139 which is not the correct answer. For instance you can draw 3 cards of the same suit a the fourth one of a different suit.
If my interpretation of your solution is not correct? Then what do you exaclty mean by "the # ways of drawing 4 cards in general". You also say "substract" something from this number (the # ways of drawing 4 cards in general), but it is not clea what to substract
• Nov 26th 2009, 07:09 AM
awkward
Quote:

Originally Posted by santiagos11
Problem 1:
There are 12 balls in an urn, 4 of which are white and 8 black. Three blndfold players, A, B, C draw a ball in turn, first A, thn B, then C. The winner is the one who first draws a white ball. Assuming that each (black) ball isreplaced after being drawn, find the ratio of the chances of the three players.

Problem 2:
There are 40 cards, 10 of each suit. A wagers B that he will draw four cards ad get one of each suit. What are the fair amount of wagers of each?

For problem 2, I know the answer (from the back of the book) is A:1000 vs. B:8139. which implies that the probability P = 1000/(1000+8139)=1000/9139. Also the number of combiantions of 4 cards out of 40 is
40 C 4 = 91390. Then the question that remains is how do i find (out of these 91390 different combinations) that there are 10^4=10000 combiantions that contain each card of a different suit?

There are 10^4 ways to choose 4 cards each of a different suit because there are 10 ways to choose the club, 10 ways to choose the spade, 10 ways to choose the heart, and 10 ways to choose the diamond.